Question

an equation of a circle is given. a determine its center. b. determine its radius c. find 2 points on the circle. x^2 + (y-8)^2= 18

Answer 1

so far, I have a. (0,8) b, approx 4.24 or sq rt 18

Answer 2

c. insert any real number for x and solve for y or the opposite

Answer 3

if x^2 > 18, you wont get the point

Answer 4

@extraven

after inserting a real no for x and getting the corresponding real no for y, what we will get, radius or a point on circle?

Answer 5

we get a point on the circle, the distance between its center and the point you get will be its radius

Answer 6

@extraven
thanks alot. i am also a beginner :) thanks alot

Answer 7

you welcome :)

Answer 8

a circle x^2 + y^2 = r^2 with centre at (0,0) and radius r

(x-a)^2 + (y-b)^2 = r^2 with centre at (positive a,positive b) and radius r

when you see x^2 you can look at it as (x-0)^2

Answer 9

why do you have to enter a number in for x that is less than 18?

Answer 10

anything (y-8) ^2 will always come out positive, since, it can either be negative squared or positive squared, both of which will be a positive

then you get x^2 + something positive = 18

the positive cannot be negative so the lowest value is 0

then it becomes x^2 + 0 = 18 and you can see why x is limited if the equation is to hold

Answer 11

ok. When I enter 3=x, I end up with sq rt 9 + 8=y.. can I simplify that to sq rt 17?